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**Thermodynamics -- PV Diagram of an Ideal Gas**

## Homework Statement

Right, I've got a PV diagram attached to make this easier to explain. The substance is an ideal gas.

I need to calculate the heat absorbed along the different paths ACB, ADB and AB. The first two are easy. The red lines in the diagram are isotherms.

## Homework Equations

PV = NKT for an ideal gas.

Q

_{AB}= ΔU + ΔW = ΔU + PΔV, U is internal energy, U = U(T) = C

_{V}T

For a monatomic ideal gas, C

_{V}= 3/2

## The Attempt at a Solution

The change in internal energy is obviously just 3/2NK(4T1 - T1) = 9/2NKT1

The work done is P

_{1}(2V

_{1}- V

_{1}) = NKT

_{1}

Thus the heat absorbed on the path AB is 11/2NKT1.

My problem is the solution manual says it's 6NKT1.

It also says that ΔU + ΔW = 3/2(NK)(2T

_{1}) + 3/2P

_{1}V

_{1}= 6NKT

_{1}which obviously isn't true so I'm questioning it's integrity. It uses the fact that Q

_{AB}= 6NKT

_{1}in the proceeding problems as well so it's a little confusing.

Thanks for any help.